# Applications of Gauss' Law - two hollow spheres

## Homework Statement

A hollow sphere of radius $r_1$ is placed at the centre of a larger hollow sphere of radius $r_2$.
Both spheres have a uniformly distributed total charge of +q
Find the preassure $p(r_2 , q)$ which acts on the outer sphere.

## Homework Equations

$\oint\textbf{E.n}dS = 4\pi k Q$

$p = \frac{dF}{dA}$

## The Attempt at a Solution

Inside the larger sphere, there is no contribution to electric field from the large sphere, so just consider the electric field at $r_2$ due to the inner sphere:

$E(4\pi r_2^2) = 4\pi k q$

$E = \frac{kq}{r_2^2}$

since $F = qE$

the force acting at $r_2$:

$F(r_2) = \frac{kq^2}{r_2^2}$

$p = \frac{F}{A}$

$p = \frac{kq^2}{4\pi r_2^4}$

does this look right?

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Just had a thought, at the point when calutating force, replace k with $\frac{1}{4\pi \epsilon_0}$

$F(r_2) = \frac{q^2}{4\pi r_2^2 \epsilon_0}$

the expression for the total area, A, of the outer sphere is now on the denominator:

$A = 4\pi r_2^2$

it follows that:

$F = \frac{q^2}{A\epsilon_0}$

now differentiate F with respect to A:

$\frac{dF}{dA} = -\frac{q^2}{A^2 \epsilon_0} = p$

$p = \frac{q^2}{16\pi^2 r_2^4 \epsilon_0}$

gthis is the same answer i got before but i guess it's a more thorough way of doing it.

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i took the center sphere to act as a point charge, and calculated E then F then P

got exactly the same. so looks good?